Given
(i) $$\text{HCN}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{CN}^-(aq)$$, $$K_a = 6.2\times 10^{-10}$$
(ii) $$\text{CN}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HCN}(aq) + \text{OH}^-(aq)$$, $$K_b = 1.6\times 10^{-5}$$.
These equilibria show the following order of the relative base strength,

The base strength of any species is measured by the equilibrium constant for the reaction in which that species accepts a proton from water.

For the cyanide ion:
$$\text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^-,$$ the equilibrium constant is its base-dissociation constant $$K_b(\text{CN}^-)=1.6\times10^{-5}.$$

For hydroxide ion the corresponding “proton-acceptance” reaction would formally be
$$\text{OH}^- + \text{H}_2\text{O} \rightleftharpoons 2\,\text{H}_2\text{O},$$ but because the product side is identical to solvent water, $$\text{OH}^-$$ is by definition the strongest base that can exist in appreciable concentration in aqueous solution. Any other base that can convert water into $$\text{OH}^-$$ is necessarily weaker than $$\text{OH}^-.$$

For water itself acting as a base:
$$\text{H}_2\text{O} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-,$$ the equilibrium constant is the ionic product of water, $$K_w = 1.0\times10^{-14}.$$ Hence water is a very weak base (large $$pK_b = 14$$).

Now compare the numerical values:

• $$K_b(\text{OH}^-) \; \text{(conceptually infinite in water)}$$
• $$K_b(\text{CN}^-) = 1.6\times10^{-5}$$
• $$K_b(\text{H}_2\text{O}) = 1.0\times10^{-14}$$

The larger the $$K_b,$$ the stronger the base. Therefore $$\text{OH}^- \; \gt \; \text{CN}^- \; \gt \; \text{H}_2\text{O}.$$

Thus the correct order of base strength is given by Option B:

Option B which is: $$\text{OH}^- \;>\; \text{CN}^- \;>\; \text{H}_2\text{O}$$