A wire having total resistance of 32 ohms is cut into 4 equal lengths. Three of them are connected in parallel and one in series to them.?

Resistance ∝ Lenght/Area

When the wire of 32 Ω is cut into 4 equal lengths :
Resistance of each length = (32 Ω) × (1/4) = 8 Ω

Equivalent resistance of the three lengths in parallel
= 1 / [(1/8) + (1/8) + (1/8)]
= 8/3 Ω
= 2.7 Ω

Equivalent resistance of combinations of 4 lengths
= (2.7 + 8) Ω
= 10.7 Ω

The resistance of each piece = 32/4 = 8 ohm
The total resistance of parallel combination = 1 /(1/8 ohm + 1/8 ohm + 1/8 ohm)
= 8/3 ohm then adding the resistance of the piece attached in series = 8/3 + 8 ohm = 32/3 ohm = 10.666 ohm

32 Ω / 4 = 8 Ω / section

R(EQ) = (1/8 Ω + 1/8 Ω + 1/8 Ω)^-1 + 8 Ω = 2.7 Ω + 8 Ω = 10.7 Ω