C

Step 1: Let the velocity of the rain be \( v_r \) and its angle with the vertical be \( \theta \). The man observes the rain at an angle of \( 60^\circ \) while stationary. The components of velocity are:
\( v_{rx} = v_r \sin(\theta) \) and \( v_{ry} = v_r \cos(\theta) \).
Step 2: When he starts moving with velocity \( v_m = 25 \, m/s \), the apparent angle of the rain becomes \( 30^\circ \). The equations become:
\( \tan(60^\circ) = \frac{v_{ry}}{v_{rx}} \) and \( \tan(30^\circ) = \frac{v_{ry}}{v_{rx} - v_m} \).
From \( \tan(60^\circ) = \sqrt{3} \), we have \( v_{ry} = \sqrt{3} v_{rx} \).
Step 3: Substituting in the second equation:
\( \tan(30^\circ) = \frac{\sqrt{3} v_{rx}}{v_{rx} - 25} \).
Simplifying, we get: \( v_{rx} = 25 \sqrt{3} \, m/s \) and \( v_{ry} = 75 \, m/s \).
Step 4: To find the velocity of rain \( v_r \), we use \( v_r = \sqrt{v_{rx}^2 + v_{ry}^2} = \sqrt{(25\sqrt{3})^2 + (75)^2} = 25\sqrt{3} \, m/s \). Therefore, the answer is option C.