At 25^ C , the concentration of H^ + ions in 1.00 × 10^ -3 M...

Because concentration of $\mathrm{H}+$ from weak acid is less we need to consider self ionization of $\mathrm{H}_2 \mathrm{O} also.$$\begin{array}{rl}\mathrm{HX}(\mathrm{aq}) \rightleftharpoons & \mathrm{H}^{+}+\mathrm{X}^{-}(\mathrm{aq}) \\ 10^{-3}-\mathrm{x} \quad & \mathrm{x}+\mathrm{y} \quad \mathrm{x} \\ \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons & \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\ & \mathrm{x}+\mathrm{y} \quad \quad \mathrm{y}\end{array}$Approximation : $\left(10^{-3}-\mathrm{x}\right) \simeq 10^{-3}\begin{aligned} & \Rightarrow \quad \frac{x(x+y)}{10^{-3}}=K_a=4 \times 10^{-11}....(1) \\ & \Rightarrow \quad y(x+y)=K w=10^{-14}.....(2) \end{aligned}\begin{array}{ll} \text{ Add }(1)+(2) \\ \end{array}\begin{array}{ll}\Rightarrow & (\mathrm{x}+\mathrm{y})^2=5 \times 10^{-14} \\ \Rightarrow & \mathrm{x}+\mathrm{y}=\left[\mathrm{H}^{+}\right]=\sqrt{5} \times 10^{-7} \\ \Rightarrow & \mathrm{x}=\sqrt{5}=2.236\end{array}$$$ Answer 2.23 or 2.24.