To solve the problem, we need to find out how long the party continued until the heights of candles A and B became equal. Let's denote:

• The length of candle A as \(L_A\).
• The length of candle B as \(L_B = 3L_A\) (since candle B is 3 times the length of candle A).
• The burning speed of candle A as \(s_A\).
• The burning speed of candle B as \(s_B = 4s_A\) (since it's burning 4 times faster than candle A).

Given, the numerical value of the length of the burnt portion of candle A is the same as its burning speed, which means after time \(t\) hours, \(L_A - s_A \cdot t = L_A - s_A\) (since it burns its own height). Therefore, the length of time for which the candles burn is given by:

\(L_A = s_A \cdot t\) \(\Rightarrow t = \frac{L_A}{s_A}\)

At the time the party ends, the heights of both candles are the same.

Equation for candle A:

\(H_A = L_A - s_A \cdot t\)

Equation for candle B:

\(H_B = L_B - s_B \cdot t\ = 3L_A - 4s_A \cdot t\)

Since both heights are equal at time \(t\):

\(L_A - s_A \cdot t = 3L_A - 4s_A \cdot t\)

Simplifying this equation:

\(L_A - s_A \cdot t = 3L_A - 4s_A \cdot t\)

Bringing like terms on one side,

\(4s_A \cdot t - s_A \cdot t = 3L_A - L_A\) \(3s_A \cdot t = 2L_A\) \(\Rightarrow t = \frac{2L_A}{3s_A}\)

From our earlier result, \(t = \frac{L_A}{s_A}\) based on candle A's consumption and time relation, gives us:

If both results should match, we equate:

\(\frac{2L_A}{3s_A} = \frac{L_A}{s_A}\)

Therefore, we can confirm the provided condition that burning speeds are based on the total length burnt throughout the time:

This results in a balanced value of time:

\(1\, \text{hour}\)

Thus, the correct answer is 1 hour.