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Given,
$\lim_{x \to 0} \frac{\sqrt{1+3x} - \sqrt{1+cx}}{cx} = 1$

Rationalizing the numerator:
$\lim_{x \to 0} \frac{(\sqrt{1+3x} - \sqrt{1+cx})(\sqrt{1+3x} + \sqrt{1+cx})}{cx(\sqrt{1+3x} + \sqrt{1+cx})} = 1$
$\lim_{x \to 0} \frac{(1+3x) - (1+cx)}{cx(\sqrt{1+3x} + \sqrt{1+cx})} = 1$
$\lim_{x \to 0} \frac{(3-c)x}{cx(\sqrt{1+3x} + \sqrt{1+cx})} = 1$

Since $x \neq 0$:
$\lim_{x \to 0} \frac{3-c}{c(\sqrt{1+3x} + \sqrt{1+cx})} = 1$
$\frac{3-c}{c(1 + 1)} = 1$
$\frac{3-c}{2c} = 1$
$3 - c = 2c$
$3c = 3 \implies c = 1$

Hence, the value of $c$ is $1$.