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$\ce{(CH_3)_2O^+ H}$ has $\rm pK_a \; -3.8$ while $\ce{H_3O^+}$ has $\rm pK_a \; 1.74\;.$ So, the former is stronger acid than latter hydronium ion.
But why is it so?
Is it due to the $\rm +I$-effect of $\ce{-(CH_3)}$? But it would rather decrease the attraction of $\ce O$ on the electron pair of $\ce H$ and that would make it indifferent to release $\ce {H^+}$ isn't it?
What is actually going that makes $\ce{(CH_3)_2O^+ H}$ stronger acid than $\ce{H_3O^+}$ ?
A powerful but often overlooked factor in determining a substance's acidity and basicity is the effect of the solvent (or lack thereof) in which it is dissolved.
You are not wrong; in the gas phase, $\ce{H_3O^{+}}$ is indeed a stronger acid than $\ce{(CH_3)_2OH^{+}}$, by a factor of about $5\times 10^{18}$. This can be seen by comparing the proton affinities of the conjugated bases, dimethyl ether ($\mathrm{804\ kJ\ mol^{-1}}$) and water ($\mathrm{697\ kJ\ mol^{-1}}$), respectively.
In other words, a mole of gaseous dimethyl ether reacting with a mole of gaseous protons releases $\mathrm{804\ kJ}$, while a mole of gaseous water releases only $\mathrm{697\ kJ}$. This means that in the gas phase, water is a weaker base than dimethyl ether, which implies that protonated water is a stronger acid than protonated dimethyl ether.
Why does the tendency invert in aqueous solution? Water can form hydrogen bonds, a very strong type of molecular interaction. $\ce{H_3O^{+}}$ has three protons available for hydrogen bond donation, whereas $\ce{(CH_3)_2OH^{+}}$ only has one. This means that the solvating water stabilises $\ce{H_3O^{+}}$ more than $\ce{(CH_3)_2OH^{+}}$. Higher stability means reduced acidity, and the difference happens to be enough that the relative order of acidity in water is the opposite of that in the gas phase.
