Me with this question. brainliest if answered.

... The top branch of the 3-branched parallel block ... the 9 and 6 in series ...
is equivalent to a single resistor of 15 ohms.

... The 3-branched parallel block boils down to (30, 10, and 15) in parallel.
That's (1/30 + 1/10 + 1/15)⁻¹  =   5 ohms.

... The 5-ohm-equivalent block and the 20-ohm resistor form a
voltage divider across the battery.
The voltage across the 5-ohm-equivalent block is (5/25 x 30v) = 6v .

... The top branch of the block is equivalent to a (9 + 6) = 15-ohmer.
With 6v across its ends, the current through that branch is (6/15) = 0.4A .

... With 0.4A flowing through it, the 9-ohm resistor is dissipating

            I²R = (0.4A)² (9 ohms)  =  (0.16 A²) (9 ohms)  =  1.44 W  (choice-3)   

Are there options for this?