At STP, a container has 1 mole of Ar, 2 moles of C O _ 2 , 3 moles...

In the first case,

${n}_{Ar} = 1, {n}_{C{O}_{2}} = 2, {n}_{{O}_{2}} = 3, {n}_{{N}_{A}} = 4$

So total number of moles $= 1 + 2 + 3 + 4 = 10$

If P in the total pressure, ${P}_{{O}_{2}} = \dfrac{3P}{10}$

In the second case,

${n}_{{O}_{2}} = 2, {n}_{total} = 9$

As P remains the same, ${P}_{{O}_{2}} = \dfrac{2P}{9}$

So decrease in partial pressure of ${O}_{2}$

$= \dfrac{3P}{10} - \dfrac{2P}{9} = \dfrac{7P}{90}$

So % decrease in partial pressure of ${O}_{2}$

$= \dfrac{\dfrac{7P}{90}}{\dfrac{3P}{10}} \times 100 = \dfrac{700}{27} = 26$