"N_0" - initial amount of U234
"N=0.25 \\times N_0" - final amount of U234
"T=2.5 \\times 10^5 \\text{ years}" - half life period
We can use the fact that
"N=0.25\\times N_0=N_0 \\Big( {1\\over 2}\\Big)^{\\frac{t}{T}} \\ \\Rightarrow 0.25= \\Big( {1\\over 2}\\Big)^{\\frac{t}{T}} , \\ \\Big( {1\\over 2}\\Big)^2=\\Big( {1\\over 2}\\Big)^{\\frac{t}{T}} \\ \\Rightarrow 2=\\frac{t}{T}"
"t=2T=5.0 \\times 10^5 \\text{ years}"
Answer: in "5\\times 10^5" years it will remain 25% of original amount.
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