Vectors

here the umbrella may be held in the direction in which the rain appears to women

so v_{rain w.r.t. women} = v_{rain w.r.t ground} -- v_{women wrt ground}

= -30k -10i

the direction in which this thing is will give you the direction and magnitude is speed with which ti hits

I didn't understand this sum.I will b grateful if u plz make me understand this sum.

The first thing to do in such problems is to reserve the vertical axes for the direction of rainfall and the horizontal axes for observer movement.
Hence we are given that the actual direction of rainfall is vertically downwards with a speed 30m/s, -30j when written in vector form.
Let the left direction be north and the right direction be south. Hence the woman is traveling in the positive direction of the X-axis, which is the positive "i" direction. Her speed is 10i m/s in vector form.

First we need the direction of rainfall relative to observer movement.

V_rw = V_r + (-V_w) = -30j - 10i m/s

We need to find the angle A.

Now to find magnitude of V_rw,
V_rw = √V²_r + V²_w + 0 as cos90° = 0.
=> V_rw = √100 + 900 = √1000
From here, you can use some easy trigonometry to find angle A. When you find it, the answer to direction will be - The woman should hold her umbrella A° south of vertical as that forward direction is south.

The speed by which rain strikes umbrella is √1000 m/s as it is the magnitude of V_rw(not sure about this one, check).

yes
firstly try to make a sketch of the situation by choosing axes...