Given,

${\left(21.5\% of 999\right)}^{\frac{1}{3}}+(43\% of 601{)}^{\frac{1}{2}}$

It can be approximated as

${\left(22\% of 1000\right)}^{\frac{1}{3}}+(\frac{40}{100}\times {600}^{}{)}^{\frac{1}{2}}=?$

 $$\begin{gathered} \begin{array}{l}\text{?= }{\left(\text{220}\right)}^{\text{1/3}}+{\left(240\right)}^{1/2}\\ \\ \text{= }{\left(\text{216}\right)}^{\text{1/3}}+{\left(225\right)}^{1/2}\\ \\ \text{= 6+15 ? }\end{array} \\ \text{= 21}\approx \text{22} \end{gathered}$$

Hence, Option B is correct.