The oxide that gives H$$_2$$O$$_2$$ most readily on treatment with H$$_2$$O is:

We begin by recalling the simple laboratory reaction that actually produces hydrogen peroxide from an oxide and water. The well-known reaction is with sodium peroxide. The balanced chemical equation is

$$\mathrm{Na_2O_2 + 2\,H_2O \;\longrightarrow\; 2\,NaOH + H_2O_2}$$

Here we can see dissociation and recombination step by step. First, water hydrates the peroxide ion, and because the peroxide ion $$\mathrm{O_2^{2-}}$$ is present in $$\mathrm{Na_2O_2}$$, one of its oxygen atoms combines with two hydrogen atoms drawn from two water molecules to form the $$\mathrm{OOH^-}$$ (hydroperoxide) intermediate; two such intermediates then couple to give one complete $$\mathrm{H_2O_2}$$ molecule, while the sodium cations are simultaneously balanced by hydroxide ions. Writing every ionic detail, we have

$$\mathrm{Na_2O_2 \; \xrightarrow{\;H_2O\;} \; 2\,Na^+ + O_2^{2-}}$$

$$\mathrm{O_2^{2-} + 2\,H_2O \;\longrightarrow\; 2\,OH^- + H_2O_2}$$

Now summing the ionic expressions, we arrive again at

$$\mathrm{Na_2O_2 + 2\,H_2O \;\longrightarrow\; 2\,NaOH + H_2O_2}$$

This reaction proceeds smoothly at room temperature, illustrating that $$\mathrm{Na_2O_2}$$ “gives” $$\mathrm{H_2O_2}$$ most readily simply by contact with water.

Let us briefly compare the other oxides mentioned:

$$\mathrm{BaO_2\cdot 8H_2O}$$ liberates $$\mathrm{H_2O_2}$$ only when treated with dilute acids, not merely with water, because the peroxide anion needs to be protonated while the barium ion must be converted to a soluble salt to shift the equilibrium.

$$\mathrm{SnO_2}$$ and $$\mathrm{PbO_2}$$ are higher oxides of tin and lead respectively. They do not contain the peroxide linkage $$\mathrm{-O-O-}$$, so when they encounter water they undergo hydrolysis to other oxides or hydroxides rather than forming $$\mathrm{H_2O_2}$$.

Since only $$\mathrm{Na_2O_2}$$ forms $$\mathrm{H_2O_2}$$ directly, it is the oxide that “gives $$\mathrm{H_2O_2}$$ most readily on treatment with $$\mathrm{H_2O}$$.”

Hence, the correct answer is Option A.