Just divide 675 by the atomic number of cesium and you will end up with
5.078798426627975
Solid KMnO₄ needed = 7.9 g
<h3>Further explanation</h3>
Given
MW KMnO₄ = 158 g/mol
500 mL(0.5 L) of a 0.1M stock solution of KMnO₄
Required
solid KMnO₄
Solution
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

Input the value :
n = M x V
n = 0.1 M x 0.5 L
n = 0.05 mol
Mass KMnO₄ :
= mol x MW
= 0.05 x 158 g/mol
= 7.9 g