To determine the value of \( R \) for which the quadratic function \( 3x^2 - 7x + R \) is a perfect square, follow these steps:

1. Rewrite the Quadratic Function:
A quadratic function \( ax^2 + bx + c \) is a perfect square if it can be expressed as \( (mx + n)^2 \). For \( 3x^2 - 7x + R \) to be a perfect square, it must be of the form:
\[
(mx + n)^2
\]

2. Expand the Perfect Square Expression:
Expand \( (mx + n)^2 \) and equate it to \( 3x^2 - 7x + R \):
\[
(mx + n)^2 = m^2x^2 + 2mnx + n^2
\]
Compare this with \( 3x^2 - 7x + R \):
\[
m^2 = 3 \\
2mn = -7 \\
n^2 = R
\]

3. Solve for \( m \) and \( n \):
- To satisfy \( m^2 = 3 \):
\[
m = \pm \sqrt{3}
\]
- Using \( 2mn = -7 \):
\[
2(\sqrt{3})n = -7 \\
n = \frac{-7}{2\sqrt{3}} = \frac{-7\sqrt{3}}{6}
\]
or
\[
2(-\sqrt{3})n = -7 \\
n = \frac{7}{2\sqrt{3}} = \frac{7\sqrt{3}}{6}
\]
- Calculate \( R \):
\[
n^2 = \left(\frac{-7\sqrt{3}}{6}\right)^2 = \frac{49 \times 3}{36} = \frac{147}{36} = \frac{49}{12}
\]
or
\[
n^2 = \left(\frac{7\sqrt{3}}{6}\right)^2 = \frac{49 \times 3}{36} = \frac{147}{36} = \frac{49}{12}
\]


The value of \( R \) that makes \( 3x^2 - 7x + R \) a perfect square is \( \frac{49}{12} \).

The correct answer is D. \( \frac{49}{12} \).