Moles of NaOH $=\frac{5 \mathrm{g}}{40 \mathrm{g}{ \mathrm{mol}}^{-1}}$= 0.125mol
Volume of the solution in litres $=450\mathrm{ml}/1000\mathrm{ml}·{\mathrm{L}}^{-1}$
$\mathrm{M}=\mathrm{n}\times \frac{1000}{\mathrm{V}\left(\mathrm{ml}\right)}$
$\text{ Molarity }=\frac{0.125\mathrm{mol}\times 1000{\mathrm{mLL}}^{-1}}{450\mathrm{mL}}$
$\begin{array}{l}=0.278\mathrm{M}=0.278{\mathrm{molL}}^{-1}\\ =0.278{\mathrm{moldm}}^{-3}\end{array}$