Pressure due to mercury column = $P = h_{\text{Hg}} \cdot \rho_{\text{Hg}} \cdot g$
Given $h_{\text{Hg}} = 0.76$ m, $\rho_{\text{Hg}} = 13.6 \times 10^3$ kg/m$^3$
Same pressure should be balanced by the new liquid column: $h_{\text{liq}} \cdot \rho_{\text{liq}} \cdot g$
So, $h_{\text{liq}} \cdot 800 = 0.76 \cdot 13600$ ⇒ $h_{\text{liq}} = \dfrac{0.76 \cdot 13600}{800} = 12.92$ m
Approximate answer = 12.9 m