1. Given standing wave equation

$y = 1.0\,\text{mm}\,\cos\left(1.57\,\text{cm}^{-1} x\right)\sin\left(78.5\,\text{s}^{-1} t\right)$

This is of the standard form

$y = A\cos(kx)\sin(\omega t)$

where

$k = 1.57\,\text{cm}^{-1}$

2. Condition for nodes

In a standing wave, nodes occur where the space-dependent factor is zero:

$\cos(kx)=0$

So,

$kx = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots$

The node closest to the origin for $x>0$ is the first one:

$kx = \frac{\pi}{2}$

3. Solve for $x$

$x = \frac{\pi/2}{1.57}$

Using $\pi \approx 3.14$,

$x = \frac{3.14/2}{1.57} = \frac{1.57}{1.57} = 1\,\text{cm}$

4. Final answer

The node closest to the origin in the region $x>0$ is at

$x = 1\,\text{cm}$