Osmotic pressure of a solution of a protein?
Ok let's start with $\text{OP} = \text{T}\times\text{R}\times\text{M}$

$\text{T} = (24.0 + 273) \text{K} = 297 \text{K}$

$\text{R} = 0.08206 \dfrac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}}$

$\text{M} = \dfrac{\text{moles}}{\text{L}} = \dfrac{\frac{37.5 \text{ g}}{1.50 \times 10^4 \text{ g/mol}}}{0.025 \text {L}} = \dfrac{2.50\times10^{-3}\text{ mol}}{0.025\text{ L}} = 0.100 \dfrac{\text{mol}}{\text{L}} $

$\text{OP} = 297\text{ K}\times0.08206 \dfrac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}}\times0.100 \dfrac{\text{mol}}{\text{L}} = 2.44 \text{ atm} $

So I get your answer too...