Answer
$γ_{}=53.9$ lbf/ft$^{3}$
Work Step by Step
We consider $1\ \mathrm{ft}^3$ of this mixture. The volume of the water in the mixture is $0.35 \mathrm{ft}^3$ which has a mass of $$
m_w=\rho_w V_w=\left(62.4 \mathrm{lbm} / \mathrm{ft}^3\right)\left(0.35 \mathrm{ft}^3\right)=21.84\ \mathrm{lbm}
$$ The weight of this water is $$
W_w=m_w g=(21.84\ \mathrm{lbm})\left(31.9\ \mathrm{ft} / \mathrm{s}^2\right)\left(\frac{1 \mathrm{lbf}}{32.174\ \mathrm{lbm} \cdot \mathrm{ft} / \mathrm{s}^2}\right)=21.65\ \mathrm{lbf}
$$ Similarly, the volume of the second fluid is $0.65\ \mathrm{ft}^3$, and the mass of this fluid is $$
m_f=\rho_f V_f=\left(50\ \mathrm{lbm} / \mathrm{ft}^3\right)\left(0.65\ \mathrm{ft}^3\right)=32.50\ \mathrm{lbm}
$$ The weight of the fluid is $$
W_f=m_f g=(32.50\ \mathrm{lbm})\left(31.9\ \mathrm{ft} / \mathrm{s}^2\right)\left(\frac{1 \mathrm{lbf}}{32.174\ \mathrm{lbm} \cdot \mathrm{ft} / \mathrm{s}^2}\right)=32.22\ \mathrm{lbf}
$$ The specific weight of this mixture is then
$$
\gamma=\frac{W_w+W_f}{V_w+V_f}=\frac{(21.65+32.22) \mathrm{lbf}}{(0.35+0.65) \mathrm{ft}^3}=\mathbf{5 3 . 9}\ \mathrm{lbf} / \mathrm{ft}^3
$$
