Hello Pavani, friend Shivam has done an excellent task. Let us congratulate that friend

But we could do the problem even simpler

*[First with 6 and 30 ohm being parallel would have 30*6/(30+6) = 5 ohm as effective*

This 5 with another 5 in series would have effective 10 ohm (top line)

In the second line we have 40 ohm which becomes parallel to 10 ohm found.

So effective will be 10*40/(10+40) = 8 ohm

This 8 ohm is in series with 2 ohm. Hence effective 10 ohm

So current coming out of battery I = V/R = 100/10 = 10 A]

This 10 A could also be found easily just by taking into account of the power dissipation in 2 ohm.

I = √P/R = √200/2 = √100 = 10 A

Now this confirms that 10 A current comes out of the battery

Ratio of resistance in top and second line is 10 : 40 i.e. 1: 4

Hence the current across them will be in the ratio 4 : 1 [recall current and resistance are inversely related]

So 10 A would be divided as 8 A through 10 ohm route and 2 A via 40 ohm route

Thus 5 ohm in the top line has 8 A passing through it

Hence power dissipation in 5 ohm resistor = I^2 * R = 8^2 * 5 = 64 * 5 = **320 W**