Some amount of a radioactive substance (half-life = 10 days ) is...

since, the initial activity is $50$~times the activity for safe occupancy, therefore ${R}_{0} = 50R,$~where $R = \lambdaN$.
since, $R ∝ N$

$\dfrac{R}{R0} = \dfrac{N}{N0} = \left(\dfrac{1}{2}\right)n = \left(\dfrac{1}{2}\right)t / T$

Or \left(\dfrac{1}{2}\right)t / 10 = \dfrac{1}{50}$</p><p>\left(2\right)t / 10 - 50 ⇒ \dfrac{t}{10}\log_{10}2 = \log_{10}50$

$⇒ t = 10\log_{10}50\log_{10}^{2} = \dfrac{10 \times 1 . 699}{0 . 301} = 56 . 4$& days