Answer
$$S G =0.775 $$
$$\gamma=7602.72 \frac{N}{m^{3}}$$
Work Step by Step
Start from definition of Specific Gravity $(\mathrm{SG}):$
$$
\begin{aligned} S G &=\frac{\rho_{f}}{\rho_{H_{2} O}} \\ S G &=\frac{775}{1000} \frac{\frac{k g}{m^{3}}}{\frac{k g}{m^{3}}} \\ S G &=0.775 \end{aligned}
$$
since $\mathrm{SG}$ in this case is an dimensionless quantity.
$$\gamma=\rho_{f} \cdot g$$
$$\gamma=775 \frac{k g}{m^{3}} \cdot 9.81 \frac{m}{s^{2}}$$
$$\gamma=7602.72 \frac{N}{m^{3}}$$
Remember, $$N=\frac{k g \cdot m}{s^{2}}$$
