Magnetic moment of A⁺³ ion is 5.48 × 10^–23 J/T (and 9.27 × 10^–24 = 1BM). If A belongs to 4th period, then element A is

Magnetic moment of A⁺³ = 5.48$\times$10^-23 J/T 

Given 1 BM = 9.27 × 10^-24 J/T

Magnetic moment of  A⁺³ = $\frac{5.48 \times {10}^{-23}}{9.27 \times {10}^{-24}}=5.9\mathrm{BM}$

$$\begin{gathered} \sqrt{\mathrm{n}(\mathrm{n}+2)}=5.9 \\ \mathrm{n}=5 \end{gathered}$$

Number of unpaired electrons in A⁺³ ion = 5

Since it belongs to 4^th period and having 5 unpaired electrons it must be a 3d -series element

Thus A⁺³ = [Ar]3d⁵

then A(z = 26) = [Ar]4s²3d⁶

Element 'A' is Iron (Fe)