For the given reaction: $Fe(OH)_3(s) \rightleftharpoons Fe^{3+}(aq) + 3OH^-(aq)$ The equilibrium constant is given by: $K_c = [Fe^{3+}][OH^-]^3$ Let the initial concentrations be $[Fe^{3+}]_1$ and $[OH^-]_1$. $K_c = [Fe^{3+}]_1 [OH^-]_1^3$ When the concentration of $OH^-$ ions is decreased to $\frac{1}{4}$ times, let the new concentration be $[OH^-]_2 = \frac{1}{4}[OH^-]_1$ and the new $Fe^{3+}$ concentration be $[Fe^{3+}]_2$. Then, $K_c = [Fe^{3+}]_2 \left(\frac{1}{4}[OH^-]_1\right)^3 = [Fe^{3+}]_2 \times \frac{1}{64} [OH^-]_1^3$ Since $K_c$ remains constant at a given temperature: $[Fe^{3+}]_1 [OH^-]_1^3 = [Fe^{3+}]_2 \times \frac{1}{64} [OH^-]_1^3$ $[Fe^{3+}]_2 = 64 \times [Fe^{3+}]_1$ Therefore, the equilibrium concentration of $Fe^{3+}$ will increase by $64$ times.