To find the value of side \( b \) in a right triangle using the Law of Cosines, given \( a = 3 \), \( c = 5 \), and \( B = 53.13^\circ \), we use the formula:
\[ b^2 = a^2 + c^2 - 2ac \cdot \cos(B) \]
Now, we can plug in the given values:
\[ b^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos(53.13^\circ) \]
Calculating \( 3^2 \) and \( 5^2 \):
\[ 3^2 = 9 \quad \text{and} \quad 5^2 = 25 \]
So,
\[ b^2 = 9 + 25 - 2 \cdot 3 \cdot 5 \cdot \cos(53.13^\circ) \]
Next, calculate \( 2 \cdot 3 \cdot 5 = 30 \), therefore:
\[ b^2 = 34 - 30 \cdot \cos(53.13^\circ) \]
Now we need to find \( \cos(53.13^\circ) \). The cosine of \( 53.13^\circ \) is approximately \( 0.6 \).
Using this value:
\[ b^2 = 34 - 30 \cdot 0.6 \]
Calculating \( 30 \cdot 0.6 = 18 \):
\[ b^2 = 34 - 18 \] \[ b^2 = 16 \]
Finally, taking the square root to find \( b \):
\[ b = \sqrt{16} = 4 \]
Rounding to the nearest whole number:
\[ b = 4 \]
So, the value of side \( b \) is \(\boxed{4}\).