Some basic facts about the Normal distribution help with this.
Background on the Normal distribution
The first fact is that any Normal random variable $Y$ with mean $\mu$ and standard deviation $\sigma$ has the same distribution as $\sigma Z + \mu$ where $Z$ is a standard Normal variable (that is, it has zero mean and unit s.d.).
The second fact is that when $(Y_1, Y_2)$ have a bivariate Normal distribution, then any linear combination $U = \alpha Y_1 + \beta Y_2$ has a Normal distribution. We may determine exactly which distribution that is by computing the mean and variance of $U$ using the usual rules,
$$E[U] = \alpha E[Y_1] + \beta E[Y_2]$$
and
$$\operatorname{Var}(U) = \alpha^2\operatorname{Var}(Y_1) + \beta^2\operatorname{Var}(Y_2) + 2\alpha\beta\operatorname{Cov}(Y_1,Y_2).$$
The third fact is that the density function of a standard Normal variable $Z$ at the value $z$ is proportional to $C\exp(-z^2/2)$ for a universal constant $C$ (whose value we don't need to know).
Because this is a density function, it integrates to unity. By a simple change of variable $z \to \alpha z + \beta$ ($\alpha \ne 0$) we can compute a host of related integrals:
$$1 = \int_{\mathbb R}C e^{-z^2/2}\,\mathrm{d}z = C\int_{\mathbb R} e^{-(\alpha z + \beta)^2/2}\,\mathrm{d}(\alpha z + \beta) = |\alpha| e^{-\beta^2/2} C \int_{\mathbb R} e^{-\alpha^2 z^2/2 - \beta z}\,\mathrm{d}z$$
which is equivalent to our fourth (and final) fact,
$$\frac{e^{\beta^2/2}}{|\alpha| } = C\int_{\mathbb R} e^{-\alpha^2 z^2/2 - \beta z}\,\mathrm{d}z.$$
Lognormal distributions
Suppose, then, that $(X_1,X_2)$ has a bivariate Normal distribution with means $\mu_i,$ standard deviations $\sigma_i,$ and covariance $\sigma_{12}=\rho\sigma_1\sigma_2$ (thus, $\rho$ is the correlation coefficient). By definition, $(X_1,X_2) = (e^{Y_1}, e^{Y_2})$ has a bivariate Lognormal distribution. Let's compute some of its moments.
The raw moments of any order $k$ are evaluated from the fourth fact as
$$E\left[X_i^k\right] = E\left[ \left(e^{Y_i}\right)^k\right] = E\left[e^{k Y_i}\right] = E\left[e^{k(\sigma_i Z_i + \mu_i)}\right] = E\left[e^{(k\sigma_i)Z_i + k\mu_i}\right] = e^{k\mu_i + (k\sigma_i)^2/2}$$
and the mixed raw moments of orders $(j,k)$ as
$$E\left[X_1^j X_2^k\right] = E\left[ \left(e^{Y_1}\right)^j \left(e^{Y_2}\right)^k\right] = E\left[e^{j Y_1 + k Y_2}\right] = e^{j\mu_1 + k\mu_2} e^{(j^2\sigma_1^2 + k^2\sigma_2^2 + 2jk\rho\sigma_1\sigma_2)/2}.$$
The last equality follows from the variance formula in the third fact, as applied to the linear combination $jY_1 + kY_2.$
Consequently, the variances and covariances are
$$S_i^2=\operatorname{Var}(X_i) = E[X_i^2] - E[X_i]^2 = e^{2\mu_i + (2\sigma_i)^2/2} - \left(e^{\mu_i + \sigma_i^2/2}\right)^2 = e^{2\mu_i + \sigma_i^2}\left(e^{\sigma_i^2}-1\right)$$
and, with similar calculations,
$$S_{12}=\operatorname{Cov}(X_1, X_2) = E[X_1X_2] - E[X_1]E[X_2] = \cdots = e^{\mu_1+\mu_2 + \sigma_1^2/2 + \sigma_2^2/2}\left(e^{\rho\sigma_1\sigma_2} - 1\right).$$
By definition, the correlation is
$$R_{12}=\operatorname{Cor}(X_1,X_2) = \frac{S_{12}}{S_1S_2} = \frac{e^{\rho\sigma_1\sigma_1} - 1}{\sqrt{(e^{\sigma_1^2} -1 )(e^{\sigma_2^2}-1)}}.$$
Answering the question
The question is tantamount to asking how to recover the covariance parameter, $\sigma_{12} = \operatorname{Cov}(Y_1,Y_2)$ in terms of the correlation and other moments of the lognormally distributed variables $(X_1,X_2).$ Writing $$M_i = E[X_i] = e^{\mu_i + \sigma_i^2/2}$$ for the expectations, easy algebra gives
$$e^{\sigma_i^2} = 1 + \frac{S_i^2}{M_i^2},$$
whence
$$\sigma_i = \sqrt{\log \left(1 + \frac{S_i^2}{M_i^2}\right)};$$
and
$$e^{\rho \sigma_1 \sigma_2} = 1 + R_{12} \frac{S_1S_2}{M_1M_2},$$
entailing
$$\sigma_{12} = \rho \sigma_1 \sigma_2 = \log\left(1 + R_{12} \frac{S_1S_2}{M_1M_2}\right).$$
The formula proposed in the question appears to be in some kind of mixed form where the Normal parameters appear on both sides. The closest I can come retains $\rho$ in the foregoing equation and re-expresses the $\sigma_i$ in terms of the moments of the $X_i$ to write
$$\sigma_{12} = \rho \sqrt{\log \left(1 + \frac{S_1^2}{M_1^2}\right)\,\log \left(1 + \frac{S_2^2}{M_2^2}\right)}.$$
