For a sparingly soluble salt that dissociates as BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq), the solubility product K_{sp} = [Ba^{2+}][SO_4^{2-}]. If the molar solubility is s mol/L, then [Ba^{2+}] = s and [SO_4^{2-}] = s, giving K_{sp} = s^2. With s = 1.05 \times 10^{-5}: K_{sp} = (1.05 \times 10^{-5})^2 = 1.1025 \times 10^{-10} \approx 1.10 \times 10^{-10}. Option 2.10 \times 10^{-5} is wrong because it is 2s rather than s^2, confusing a simple multiplication with a squaring operation. Option 1.05 \times 10^{-5} is the solubility s itself, not K_{sp}. Option 4.41 \times 10^{-25} would come from cubing s (appropriate for salts dissociating into 3 ions), not squaring, which is the correct operation for a 1:1 electrolyte. This K_{sp} calculation is a fundamental NCERT ionic equilibrium problem and appears regularly in JEE Main. Plausibility check: the known K_{sp} of BaSO_4 is approximately 1.1 \times 10^{-10}, confirming our calculated value.