1. Let the rain's velocity relative to the ground have components:

   • Horizontal component = $v_h$
   • Vertical downward component = $v_v$

2. First situation: girl is standing still

   She holds the umbrella at $45^\circ$ with the vertical, so the rain appears to come at $45^\circ$ with the vertical.

   Hence, $\tan 45^\circ = \frac{v_h}{v_v} = 1$ so, $v_h = v_v$

3. Second situation: girl runs without umbrella

   Her speed is $15\sqrt{2}\ \text{km h}^{-1}$

   Now the rain hits her vertically, which means the horizontal component of rain relative to the girl becomes zero.

   Therefore, girl's horizontal speed must equal the horizontal speed of rain: $v_h = 15\sqrt{2}$

   From step 2, $v_v = 15\sqrt{2}$

4. Velocity of rain relative to the moving girl

   Since the relative horizontal component is zero, only the vertical component remains.

   Thus relative speed of rain with respect to the girl is $v_{rg} = v_v = 15\sqrt{2}$

   Rationalizing/comparing with options: $15\sqrt{2} = \frac{30}{\sqrt{2}}\ \text{km h}^{-1}$

5. Match with options

   $\frac{30}{\sqrt{2}}\ \text{km h}^{-1}$ corresponds to Option C.

Verification with stored answer: Stored correct answer is C, which matches the derived answer.