find the resistance?

The power all goes to heat energy in the resistors.  The easiest way to get an answer is to find the equivalent series resistance r, and then the power is P = i²r, where i is the 2A current provided by the source.

The circuit has resistance of R+5 and 10 in parallel, so:

   1/r = 1/(R+5) + 1/10 = (R + 15)/(10R + 50)
   r = (10R + 50)/(R + 15)

   i²r = 24
   4 (10R + 50) / (R + 15) = 24
    10R + 50 = 6(R + 15)
    4R = 90 - 50 = 40
    R = 10

Let R' be the equivalent resistance of the 5 Ω resistor and R.
Let I A be the current passing through the 5 Ω resistor.
Then, the current passing through the 10 Ω resistor = (2 - I) A

Voltage (in V) across the 2 A source:
(2 - I) × 10 = I × R' …… [1]

Total power (in W):
(2 - I)² × 10 + I² × R' = 24
(2 - I) × [(2 - I) × 10] + I² × R'= 24 …… [2]

Plug [1] into [2]:
(2 - I) × [I × R'] + I² × R' = 24
2I(5 + R) - I²R' + I²R' = 24
2IR' = 24
I = 12/R' …… [3]

Plug [3] into [1]:
[2 - (12/R')] × 10 = (12/R') × R'
20 - (120/R') = 12
120/R' = 8
R' = 15

R' = R + 5
15 = R + 5
R = 10

The answer: (c) 10 Ω

Using P = VI, we have 24 = 2V

i.e. V = 12

In the right loop we have 12 - 10I₁ = 0

so, I₁ = 6/5 => 1.2 Amps

Hence, current through the left loop is 0.8 Amps

so, 12 - 5(0.8) - 0.8R = 0

=> 8 = 0.8R

so, R = 10 Ω

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