Concentration of solution $ =0.1$

Degree of ionisation $ = 2\% = \frac{2}{{100}} = 0.02$

Ionic product of water $ = 1 \times {10^{ - 14}}$

Concentration of $[{H^ + }]$= Concentration of solution $\times$ degree of ionisation $ = 0.1 \times 0.02 = 2 \times {10^{ - 3}}M$

Concentration of $[O{H^ - }] = \frac{{{\rm{Ionic product of water}}}}{{[{H^ + }]}}$

$ = \frac{{1 \times {{10}^{ - 14}}}}{{2 \times {{10}^{ - 3}}}} = 0.5 \times {10^{ - 11}} = 5 \times {10^{ - 12}}M$.