Given:
Width, \( b = 250 { mm} \)
Effective depth, \( d = 500 { mm} \)
Design shear strength, \( \tau_c = 0.3 { N/mm}^2 \)
Step 1: Calculate shear force,
\[
V_e = \tau_c \cdot b \cdot d = 0.3 \times 250 \times \frac{500}{1000} = 37.5\ {kN}
\]
Step 2: Use IS 456:2000 torsion formula,
\[
T_u = \frac{V_e \cdot b}{1.6 + \frac{1.67 b}{d}} = \frac{37.5 \times 0.25}{1.6} = 5.86\ {kN.m}
\]
Hence, the torsional moment capacity is \(\boxed{5.86\ {kN.m}}\).