Resistance of a wire, $$R = {{\rho l} \over A}$$

For iron (Fe) bar,

$$\rho$$ = 10^$$-$$7 $$\Omega$$m, l = 50 mm = 50 $$\times$$ 10^$$-$$3 m

A = (2 mm) $$\times$$ (2 mm) = 4 mm² = 4 $$\times$$ 10^$$-$$6 m²

$${R_1} = {{{{10}^{ - 7}} \times 50 \times {{10}^{ - 3}}} \over {4 \times {{10}^{ - 6}}}} = 1250 \times {10^{ - 6}}$$ $$\Omega$$ = 1250 $$\mu$$W

For aluminium (Al) bar,

$$\rho$$ = 2.7 $$\times$$ 10^$$-$$8 $$\Omega$$ m, l = 50 mm = 50 $$\times$$ 10^$$-$$3 m

A = (7² $$-$$ 2²) mm² = 45 mm² = 45 $$\times$$ 10^$$-$$6 m²

$$\therefore$$ $${R_2} = {{2.7 \times {{10}^{ - 8}} \times 50 \times {{10}^{ - 3}}} \over {45 \times {{10}^{ - 6}}}}$$

$$ = {{27 \times 50} \over {45}} \times {10^{ - 6}} = 30 \times {10^{ - 6}}$$ $$\Omega$$ = 30 $$\mu$$$$\Omega$$

Potential difference across both bars (resistors) is same so they are in parallel combination.

Equivalent resistance between P and Q is given by

$$R = {{{R_1}{R_2}} \over {{R_1} + {R_2}}} = {{1250 \times 30} \over {1250 + 30}} = {{125 \times 30} \over {128}} = {{1875} \over {64}}\mu \Omega $$.