1. Use the midpoint form of a chord of a conic

For the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1,$ a chord whose midpoint is $(x_1,y_1)$ is given by the standard formula $T=S_1,$ where

• $S\equiv \frac{x^2}{25}+\frac{y^2}{16}-1,$
• $T\equiv \frac{xx_1}{25}+\frac{yy_1}{16}-1,$
• $S_1\equiv \frac{x_1^2}{25}+\frac{y_1^2}{16}-1.$

Here, the midpoint is $(3,1)$.

2. Compute $S_1$ at $(3,1)$

$S_1=\frac{3^2}{25}+\frac{1^2}{16}-1=\frac{9}{25}+\frac{1}{16}-1.$ Taking LCM $400$, $\frac{9}{25}=\frac{144}{400},\qquad \frac{1}{16}=\frac{25}{400}.$ So, $S_1=\frac{144+25}{400}-1=\frac{169}{400}-1=-\frac{231}{400}.$

3. Write $T=S_1$

$\frac{3x}{25}+\frac{y}{16}-1=-\frac{231}{400}.$ Move $-1$ to the right side: $\frac{3x}{25}+\frac{y}{16}=1-\frac{231}{400}=\frac{169}{400}.$

4. Clear denominators

Multiply by $400$: $400\cdot \frac{3x}{25}+400\cdot \frac{y}{16}=169.$ That gives $48x+25y=169.$

5. Match with the options

This is Option B.

Therefore, the equation of the chord is $\boxed{48x+25y=169}.$