Dipole moment= $\mathrm{q}\times \mathrm{d}$

Bond length $\mathrm{d}=150\mathrm{pm}=150\times {10}^{-12}\mathrm{m}$

Charge $\mathrm{q}=1.6\times {10}^{-19}$

Dipole moment = $150\times {10}^{-12}\times 1.6\times {10}^{-19}=2.4\times {10}^{-29}$

Percentage ionic character =$\frac{\mathrm{Dipole} \mathrm{observed}}{\mathrm{Dipole} \mathrm{calculated}}\times 100$

Dipole observed = $1.5\times {10}^{-29}$

Percentage ionic character = $\frac{1.5\times {10}^{-29}}{2.4\times {10}^{-29}}\times 100=62.5$

Hence, the percent ionic character of the bond will be $62.5\%$.