Answer
$$3.1 \times 10^{31} \mathrm{m}^{-3}
$$
Work Step by Step
From $\rho_{\mathrm{H}}=0.35 \rho=n_{p} m_{p}$, we get the proton number density $n_{p}:$
$$
n_{p}=\frac{0.35 \rho}{m_{p}}=\frac{(0.35)\left(1.5 \times 10^{5} \mathrm{kg} / \mathrm{m}^{3}\right)}{1.67 \times 10^{-27} \mathrm{kg}}=3.1 \times 10^{31} \mathrm{m}^{-3} .
$$
