Decreasing order (first having highest and then others following it) of mass of pure NaOH in each of the aqueous solution:
(i) 50 g of 40% (w/W) NaOH
(ii) 50 mL of 50% (w/V) NaOH [d_soln. = 7.2 g / mL]
(iii) 50 g of 15 M NaOH [d_soln. : 1g,/mL]

(i) Mass of NaOH $=50\times \frac{40}{100}=20\mathrm{g}$
(ii) $\frac{\mathrm{w}}{\mathrm{v}}\mathrm{NaOH}\to 50 \mathrm{g} \mathrm{NaOH} \mathrm{in} 100\mathrm{mL}$
So, ln 50 mL mass of NaOH will be =25 g

(iii)  50 g of 15 M NaOH(d =1g/mL)
no.of moles n = MV = $\frac{\mathrm{W}}{\mathrm{GMW}}$

                  $$\begin{gathered} \mathrm{W} =50\times 15\times 40 \\ =300 \end{gathered}$$                               

Order (iii) > (ii) > (i)