Correct option is (4) 4Ω
All the wires are identical and of same material, so they will have the same value of resistance.
Let it be R, when these are (four) connected in parallel,

\(R_P = \frac R4\) \(\left(\because \frac 1{R_p} = \frac{1}{R_1} +\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4} \right)\)
Given, Rp = 0.25 Ω
∴ 0.25 = \(\frac R4\)
∴ R = 1Ω
Now, these four resistances are arranged in series

Rs = R + R + R + R
= 4R
= 4 × 1
= 4Ω